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A student is at a distance $16 \mathrm{~m}$ from a bus when the bus begins to move with a constant acceleration of $9 \mathrm{~m} \mathrm{~s}^{-2}$. The minimum velocity with which the student should run towards the bus so as the catch it is $\alpha \sqrt{2} \mathrm{~ms}^{-1}$. The value of $\alpha$ is
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$12$
Let $v$ be the minimum velocity of student so,that he could catch the bus
If student catch the bus in time $t$, then distance travelled by student in time $t$ $=16+$ distance travelled by bus in time $t$.
$\begin{aligned} & \Rightarrow v t=16+\left(u t+\frac{1}{2} a t^2\right) \\ & \Rightarrow v t=16+0 \times t+\frac{1}{2} \times 9 \times t^2 \\ & \Rightarrow v t=16+\frac{9}{2} t^2 \Rightarrow 9 t^2-2 v t+32=0\end{aligned}$
The above equation must have real roots. i.e its discriminant $\geq 0$
i.e $(2 v)^2-4 \times 9 \times 32 \geq 0$
$\begin{aligned} & \Rightarrow 4 v^2-4 \times 288 \geq 0 \Rightarrow v^2-288 \geq 0 \\ & \Rightarrow v^2 \geq 288\end{aligned}$
$v \geq 12 \sqrt{2} \mathrm{~m} / \mathrm{s}$
Minimum velocity of student to catch the bus $=12 \sqrt{2} \mathrm{~m} / \mathrm{s}=\alpha \sqrt{2} \mathrm{~m} / \mathrm{s}$ (given)
$\therefore \alpha=12$
If student catch the bus in time $t$, then distance travelled by student in time $t$ $=16+$ distance travelled by bus in time $t$.
$\begin{aligned} & \Rightarrow v t=16+\left(u t+\frac{1}{2} a t^2\right) \\ & \Rightarrow v t=16+0 \times t+\frac{1}{2} \times 9 \times t^2 \\ & \Rightarrow v t=16+\frac{9}{2} t^2 \Rightarrow 9 t^2-2 v t+32=0\end{aligned}$
The above equation must have real roots. i.e its discriminant $\geq 0$
i.e $(2 v)^2-4 \times 9 \times 32 \geq 0$
$\begin{aligned} & \Rightarrow 4 v^2-4 \times 288 \geq 0 \Rightarrow v^2-288 \geq 0 \\ & \Rightarrow v^2 \geq 288\end{aligned}$
$v \geq 12 \sqrt{2} \mathrm{~m} / \mathrm{s}$
Minimum velocity of student to catch the bus $=12 \sqrt{2} \mathrm{~m} / \mathrm{s}=\alpha \sqrt{2} \mathrm{~m} / \mathrm{s}$ (given)
$\therefore \alpha=12$
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